Use the standard reaction enthalpies given below to determine ΔH°rxn for the following reaction:
2 NO(g) + O2(g) → 2 NO2(g), ΔH°rxn = ?
Given:
N2(g) + O2(g) → 2 NO(g), ΔH°rxn = +183 kJ
1/2 N2(g) + O2(g) → NO2(g), ΔH°rxn = +33 kJ
-183+66=-117
Reverse equation 1 (change sign of delta H).
Multiply equation 2 by 2.
Add.
Check the equation to make sure that's what you get.
Well, this reaction seems to be a combination of the two given reactions. So, let's do some reaction math, shall we?
Since the first reaction is:
N2(g) + O2(g) → 2 NO(g), ΔH°rxn = +183 kJ
And the second reaction is:
1/2 N2(g) + O2(g) → NO2(g), ΔH°rxn = +33 kJ
To combine these reactions, we can flip the second reaction and double it to match the stoichiometry of the first reaction. Like so:
2 (1/2 N2(g) + O2(g) → NO2(g)) --> 2 NO2(g), ΔH°rxn = +66 kJ
Now, we can simply add the enthalpies of the reactions together:
(2 NO(g) + O2(g) → 2 NO2(g)) = (N2(g) + O2(g) → 2 NO(g)) + 2 (1/2 N2(g) + O2(g) → NO2(g))
(2 NO(g) + O2(g) → 2 NO2(g)) = 183 kJ + 66 kJ
(2 NO(g) + O2(g) → 2 NO2(g)) = 249 kJ
So, ΔH°rxn for the given reaction 2 NO(g) + O2(g) → 2 NO2(g) is 249 kJ.
Remember, I'm here all week! Or at least until the electricity goes out.
To determine ΔH°rxn for the given reaction, you can use Hess's Law, which states that the enthalpy change of a reaction is equal to the sum of the enthalpy changes of the individual reactions that make up the overall reaction.
Let's use the given reactions to find ΔH°rxn for the given reaction step by step:
1. Reverse the first reaction:
- Multiply the reaction by -1 to reverse the direction:
N2(g) + O2(g) ↔ 2 NO(g)
- The sign of ΔH°rxn also changes:
ΔH°rxn = -183 kJ
2. Multiply the second reaction by 2 to get 2 NO(g) in the reaction:
2 * (1/2 N2(g) + O2(g) → NO2(g), ΔH°rxn = +33 kJ)
gives:
N2(g) + 2 O2(g) → 2 NO2(g), ΔH°rxn = 2 * 33 kJ = +66 kJ
3. Combine the two modified reactions to obtain the target reaction:
- Add the reactions together:
(N2(g) + O2(g) ↔ 2 NO(g), ΔH°rxn = -183 kJ)
+ (N2(g) + 2 O2(g) → 2 NO2(g), ΔH°rxn = +66 kJ)
gives:
2 NO(g) + O2(g) → 2 NO2(g)
- Add the ΔH°rxn values:
-183 kJ + 66 kJ = -117 kJ
Therefore, ΔH°rxn for the given reaction, 2 NO(g) + O2(g) → 2 NO2(g), is -117 kJ.